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State whether each of the following series converges absolutely, conditionally, or not at all\(\sum_{n=1}^{\infty}(-1)^{n+1} \sin ^{2}(1 / n)\)

Short Answer

Expert verified
The series converges absolutely.

Step by step solution

01

Identify the Series Type

The given series is \[ \sum_{n=1}^{\infty}(-1)^{n+1} \sin ^{2}\left(\frac{1}{n}\right) \]. This is an alternating series because of the factor \((-1)^{n+1}\) that changes sign for each subsequent term.
02

Test for Absolute Convergence

To check for absolute convergence, consider \[ \sum_{n=1}^{\infty} \left| (-1)^{n+1} \sin ^{2}\left(\frac{1}{n}\right) \right| = \sum_{n=1}^{\infty} \sin ^{2}\left(\frac{1}{n}\right).\]For large \(n\), \(\sin\left(\frac{1}{n}\right) \approx \frac{1}{n}\), so \(\sin^2\left(\frac{1}{n}\right) \approx \left(\frac{1}{n}\right)^2 = \frac{1}{n^2}\). The series \(\sum_{n=1}^{\infty} \frac{1}{n^2}\) is known to converge, thus this series converges absolutely.
03

Conclusion on Convergence

Since the series \(\sum_{n=1}^{\infty} \left| (-1)^{n+1} \sin ^{2}\left(\frac{1}{n}\right) \right|\) converges, the original series \(\sum_{n=1}^{\infty}(-1)^{n+1} \sin ^{2}\left(\frac{1}{n}\right)\) converges absolutely.There's no need to test for conditional convergence once absolute convergence is established.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Alternating Series
An alternating series is a specific type of series where the signs of the terms alternate between positive and negative. This happens due to a factor like \((-1)^{n+1}\), which switches signs based on the values of \(n\).
Alternating series appear frequently in mathematical problems, especially when dealing with oscillating processes. These series exhibit a distinct pattern which can be easier to analyze compared to other complex series.
One of the well-known tests for these types of series is the Alternating Series Test. For a series \(\sum_{n=1}^{\infty} (-1)^{n+1} b_n\), it converges if two conditions are met:
  • The absolute value of terms \(b_n\) decreases (i.e., \(b_{n+1} \leq b_n\) for all \(n\)).
  • The limit of \(b_n\) as \(n\) approaches infinity is zero \((\lim_{{n \to \infty}} b_n = 0)\).
This test allows us to conclude the convergence of many series without needing complex computations.
However, to move from determining convergence to establishing the nature of that convergence (absolute or conditional) requires further analysis.
Absolute Convergence
In mathematics, when we discuss series and convergence, the term "absolute convergence" is significant. A series \(\sum a_n\) converges absolutely if the series of absolute values \(\sum |a_n|\) also converges.
Absolute convergence is a stronger form of convergence compared to simple convergence. If a series converges absolutely, it also converges conditionally, but not necessarily vice versa.
Knowing that a given series \(\sum (-1)^{n+1} \sin^2\left(\frac{1}{n}\right)\) converges absolutely because \(\sum \sin^2\left(\frac{1}{n}\right)\) behaves like \(\sum \frac{1}{n^2}\), which is a known convergent series, strengthens our understanding.
It's important to remember that once absolute convergence is verified for a series, there's no need to check for conditional convergence, as this ensures both types exist.
Series Approximation
Series approximations are incredibly useful in understanding the behavior of functions, especially when they become difficult to compute directly. Approximations involve replacing complex infinite processes with simpler expressions that are easy to handle.
In the context of determining convergence like in our example series, spotting approximations can simplify the problem. Here, \(\sin^2\left(\frac{1}{n}\right)\) can be approximated using \(\left(\frac{1}{n}\right)^2\) for large \(n\), since \(\sin\left(\frac{1}{n}\right) \approx \frac{1}{n}\) when \(n\) is large.
This approximation supports determining whether the series converges, as \(\sum \frac{1}{n^2}\) already denotes a convergent p-series \((\text{where } p > 1)\).
Approximations are powerful because they help tackle infinite behaviors with finite understanding, making complex series manageable and their properties easier to comprehend.
Convergence Criteria
Understanding convergence criteria is your tool kit for analyzing whether a series like \(\sum (-1)^{n+1} \sin^2\left(\frac{1}{n}\right)\) converges. Various criteria aid in verifying the nature of convergence.
A key one, as mentioned, is the Absolute Convergence Test, where the convergence of \(\sum |a_n|\) implies the convergence of \(\sum a_n\).
Additionally, there's Conditional Convergence. If \(\sum a_n\) converges, but \(\sum |a_n|\) does not, it's conditionally convergent. This matters in fields like numerical computation and series manipulation.
Understanding these criteria aids in distinguishing series behaviors, especially when determining which convergence form is present. Each test offers insights that help to predict and verify convergence in mathematical series.
Always begin with an overview of the series structure, apply appropriate tests, and certainly look for approximations or simplifications that make analysis straightforward.

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Most popular questions from this chapter

True or False? Justify your answer with a proof or a counterexample. $$ \text { If } \lim _{n \rightarrow \infty} a_{n}=0, \text { then } \sum_{n=1}^{\infty} a_{n} \text { converges } $$

Is the sequence bounded, monotone, and convergent or divergent? If it is convergent, find the limit. $$ a_{n}=\frac{\ln (\cos n)}{n} $$

Use the ratio test to determine whether \(\sum_{n=1}^{\infty} a_{n}\) converges, where \(a_{n}\) is given in the following problems. State if the ratio test is inconclusive. $$ \sum_{n=1}^{\infty} \frac{\left(2^{n} n !\right)^{2}}{(2 n)^{2 n}} $$

The following advanced exercises use a generalized ratio test to determine convergence of some series that arise in particular applications when tests in this chapter, including the ratio and root test, are not powerful enough to determine their convergence. The test states that if \(\lim _{n \rightarrow \infty} \frac{a_{2 n}}{a_{n}}<1 / 2\), then \(\sum a_{n}\) converges, while if \(\lim _{n \rightarrow \infty} \frac{a_{2 n+1}}{a_{n}}>1 / 2\), then \(\sum a_{n}\) diverges. Let \(a_{n}=\frac{1}{1+x} \frac{2}{2+x} \cdots \frac{n}{n+x} \frac{1}{n}=\frac{(n-1) !}{(1+x)(2+x) \cdots(n+x)} .\) Show that \(a_{2 n} / a_{n} \leq e^{-x / 2} / 2 .\) For which \(x>0\) does the generalized ratio test imply convergence of \(\sum_{n=1}^{\infty} a_{n} ?\) (Hint: Write \(2 a_{2 n} / a_{n}\) as a product of \(n\) factors each smaller than \(1 /(1+x /(2 n))\)

Suppose that \(\sum_{n=1}^{\infty} a_{n}\) is a convergent series of positive terms. Explain why \(\lim _{N \rightarrow \infty} \sum_{n=N+1}^{\infty} a_{n}=0\).

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